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Semiprimes and Emirps (Posted on 2023-01-14) Difficulty: 3 of 5
Determine the total number of semiprimes below 2023, that are immediately followed by an emirp.

For example, 22 is a semiprime which is immediately followed by 23. But 23 is NOT an emirp, since 32 is composite.

List all valid pairs in conformity with the provisions of the problem.

*** Computer program/ excel solver aided methods are welcome, but a semi-analytic methodology is preferred.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
semi-analytic solution | Comment 2 of 5 |
Assumptions: 

The prime number theorem says that pp(y), the probability that y is a prime ~ 1/ln(y)

Ishmukhametov and Sharifullina say that ps(y), the probability that y is a semi-prime ~ ln[ln(y)] / ln(y) + A/ln(y) + B/[ln(y)]^2
with A = 0.26 and B = -1.54

I say that probability of an emirp ~ pp^2 (to have it both ways) 
This is a very rough approximation that depends on the fact that a number reversed is of the same order of magnitude (has the same number of digits).  For 4 digits, pp ranges only from 0.1 to 0.14, 
and further, pp^2 is equally likely over- and under-estimates pp*pr, where pr is the probability that y reversed is prime.  But, pp^2 is also an over-estimate as it includes palindromic primes (which are not emirps). 
 
I say the probability of y semi-prime followed by y+1 emirp is
 ~~ ps(y) * pp(y+1)^2.

So,
sum(y=13 to 2023) [ps pp^2] = 14.5

note: Larry found 12 in this range (13-2023) (including the palindromes)
  program math
sum=0
   do i=13,2023
   y=i
   x=log(y)
   pp=1/log(y+1)
   ps=log(x)/x+0.265/x-1.54/x**2
   sum=sum+pp**2*ps
   enddo
print*,sum
end
>math
14.5202

Edited on January 15, 2023, 3:58 am
  Posted by Steven Lord on 2023-01-14 10:54:39

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